Question

A stone is thrown at $25 \, \text{m} / \text{s}$ at $53^{o}$ above the horizontal. At what time its velocity is at $45^{o}$ below the horizontal?

• A. $0.5 \, \, \text{s}$
• B. $4 \, \, \text{s}$
• C. $3.5 \, \, \text{s}$
• D. $2.5 \, \, \text{s}$

Answer 1

The Correct Option is C

Horizontal component of velocity, throughout the motion remain constant using $ucos \, \theta =vcos \, \alpha $ $25cos 53 ^\circ = v cos ? 45 ^\circ $ $\Longrightarrow v=25\times \frac{3}{5}.\sqrt{2}$ $=15\sqrt{2}$ $\Longrightarrow v_{y}=vsin 45 = 15$ Now using $ \, \, \, V_{y}=u_{y}+a_{y}t \, in \, \, y-$ direction, $-15=25sin 53? $ $-15=25\times \frac{4}{5}-10 \, t$ $\Longrightarrow t=3.5 \, \text{sec}$